# Mathematics 2011 | Study Mode

Question 11
A
2X - 1
B
X + 3
C
2X + 1
D
X - 3
##### Explanation
X2 + 1 $$\frac{X - 2}{\sqrt{X^3 - 2X^2 + 3n - 3}}$$

= $$\frac {- 6X^3 + n}{-2X^2 + 2X - 3}$$

= $$\frac{(-2X^2 - 2)}{2X - 1}$$

Remainder is 2X - 1

Question 12
A
(3y - 2x)(3y + 4x)
B
(3y + 4x)(3y + 4x)
C
(3y + 2x)(3y - 4x)
D
(3y - 4x)(3y + 4x)
##### Explanation
9y2 - 16x2

= 32y2 - 42x2

= (3y - 4x)(3y +4x)

Question 13
A
-3, -9
B
9, -3
C
-9,3
D
3, -9
##### Explanation
-2x -5y = 3

x + 3y = 0

x = -3y

-2 (-3y) - 5y = -3

6y - 5y = 3

y = 3

but, x = -3y

x = -3(3)

x = -9

therefore, x = -9, y = 3

Question 14
A
20$$\frac{1}{4}$$
B
27
C
2$$\frac{1}{4}$$
D
36
##### Explanation
x $$\alpha\sqrt y$$

x = k$$\sqrt y$$

81 = k$$\sqrt9$$

k = $$\frac{81}{3}$$

= 27

therefore, x = 27$$\sqrt y$$

y = 1$$\frac{7}{9}$$ = $$\frac{16}{9}$$

x = 27 x $$\sqrt{\frac{16}{9}}$$

= 27 x $$\frac{4}{3}$$

dividing 27 by 3

= 9 x 4

= 36

Question 15
A
$$\frac{1}{18}$$
B
$$\frac{1}{12}$$
C
$$\frac{1}{24}$$
D
$$\frac{1}{6}$$
##### Explanation
T $$\alpha \frac{1}{R^3}$$

T = $$\frac{k}{R^3}$$

k = TR3

= $$\frac{2}{81}$$ x 33

= $$\frac{2}{81}$$ x 27

dividing 81 by 27

k = $$\frac{2}{2}$$

therefore, T = $$\frac{2}{3}$$ x $$\frac{1}{R^3}$$

When R = 2

T = $$\frac{2}{3}$$ x $$\frac{1}{2^3}$$ = $$\frac{2}{3}$$ x $$\frac{1}{8}$$

= $$\frac{1}{12}$$

Question 16
A
x $$\leq$$ 2
B
x $$\geq$$ -1
C
x $$\geq$$ -2
D
x $$\leq$$ -1
##### Explanation
-6(x + 3) $$\leq$$ 4(x - 2)

-6(x +3) $$\leq$$ 4(x - 2)

-6x -18 $$\leq$$ 4x - 8

-18 + 8 $$\leq$$ 4x +6x

-10x $$\leq$$ 10x

10x $$\leq$$ -10

x $$\leq$$ 1

Question 17
A
x < -3 or x > 5
B
-5 < x < 3
C
x < 3 or x > 5
D
x > 3 or x < -5
##### Explanation
x2 + 2x > 15

x2 + 2x - 15 > 0

(x2 + 5x) - (3x - 15) > 0

x(x + 5) - 3(x + 5) >0

(x - 3)(x + 5) > 0

therefore, x = 3 or -5

then x < -5 or x > 3

i.e. x< 3 or x < -5

Question 18
A
505
B
513
C
433
D
635
##### Explanation
3, 6, 9,..., 36.

a = 3, d = 3, i = 36, n = 18

Sn = $$\frac{n}{2}$$ [2a + (n - 1)d

S18 = $$\frac{18}{2}$$ [2 x 3 + (18 - 1)3]

= 9[6 + (17 x 3)]

= 9 [6 + 51] = 9(57)

= 513

Question 19
A
60
B
62
C
54
D
64
##### Explanation
T2 = 4, T4 = 16

Tx = arn-1

T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4

T4 = ar4-1

therefore, $$\frac{T_4}{T_r}$$ = $$\frac{ar^3}{ar}$$ = $$\frac{16}{4}$$

r2 = 4 and r = 2

but ar = 4

a = $$\frac{4}{r}$$ = $$\frac{4}{2}$$

a = 2

Sn = $$\frac{a(r^n - 1)}{r - 1}$$

S5 = $$\frac{2(2^5 - 1)}{2 - 1}$$

= $$\frac{2(32 - 1)}{2 - 1}$$

= 2(31)

= 62

Question 20
A
- $$\frac{1}{2}$$
B
$$\frac{1}{3}$$
C
-1
D
2
##### Explanation
N + Y = XY + X + Y

3 + -$$\frac{2}{3}$$ = 3(- $$\frac{2}{3}$$) + 3 + (- $$\frac{2}{3}$$)

= -2 + 3 -$$\frac{2}{3}$$

= $$\frac{1 - 2}{1 - 3}$$

= $$\frac{1}{3}$$

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